Simplify and expand the following expression: $ \dfrac{3}{4n + 8}- \dfrac{1}{n - 7}+ \dfrac{4}{n^2 - 5n - 14} $
Answer: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $4$ out of denominator in the first term: $ \dfrac{3}{4n + 8} = \dfrac{3}{4(n + 2)}$ We can factor the quadratic in the third term: $ \dfrac{4}{n^2 - 5n - 14} = \dfrac{4}{(n + 2)(n - 7)}$ Now we have: $ \dfrac{3}{4(n + 2)}- \dfrac{1}{n - 7}+ \dfrac{4}{(n + 2)(n - 7)} $ The least common multiple of the denominators is: $ 4(n + 2)(n - 7)$ In order to get the first term over $4(n + 2)(n - 7)$ , multiply by $\dfrac{n - 7}{n - 7}$ $ \dfrac{3}{4(n + 2)} \times \dfrac{n - 7}{n - 7} = \dfrac{3(n - 7)}{4(n + 2)(n - 7)} $ In order to get the second term over $4(n + 2)(n - 7)$ , multiply by $\dfrac{4(n + 2)}{4(n + 2)}$ $ \dfrac{1}{n - 7} \times \dfrac{4(n + 2)}{4(n + 2)} = \dfrac{4(n + 2)}{4(n + 2)(n - 7)} $ In order to get the third term over $4(n + 2)(n - 7)$ , multiply by $\dfrac{4}{4}$ $ \dfrac{4}{(n + 2)(n - 7)} \times \dfrac{4}{4} = \dfrac{16}{4(n + 2)(n - 7)} $ Now we have: $ \dfrac{3(n - 7)}{4(n + 2)(n - 7)} - \dfrac{4(n + 2)}{4(n + 2)(n - 7)} + \dfrac{16}{4(n + 2)(n - 7)} $ $ = \dfrac{ 3(n - 7) - 4(n + 2) + 16} {4(n + 2)(n - 7)} $ Expand: $ = \dfrac{3n - 21 - 4n - 8 + 16}{4n^2 - 20n - 56} $ $ = \dfrac{-n - 13}{4n^2 - 20n - 56}$